Find the equation of tangent to the curve y=2x-x^3 at the point x=-1. Where does that tangent meet the line y=x?
To find the equation of the tangent to the curve y = 2x - x^3 at the point x = -1, we need to find the slope of the tangent line and the coordinates of the point where it meets the line y = x.
First, let's find the derivative of y with respect to x to obtain the slope of the tangent line:
y = 2x - x^3
dy/dx = 2 - 3x^2
Next, substitute x = -1 into dy/dx to find the slope:
dy/dx = 2 - 3(-1)^2
dy/dx = 2 - 3(1)
dy/dx = 2 - 3
dy/dx = -1
The slope of the tangent line to the curve at x = -1 is -1.
Now, let's find the y-coordinate of the point where the tangent line meets the line y = x. We can substitute the x-coordinate of the point (-1) into the original equation y = 2x - x^3:
y = 2(-1) - (-1)^3
y = -2 - (-1)
y = -2 + 1
y = -1
The y-coordinate of the point where the tangent line meets the line y = x is -1.
Therefore, the tangent line has a slope of -1 and passes through the point (-1, -1). Using the point-slope form of a linear equation, we can write the equation of the tangent line:
y - (-1) = -1(x - (-1))
y + 1 = -1(x + 1)
y + 1 = -x - 1
y = -x - 2
So, the equation of the tangent line is y = -x - 2.
To find the point where the tangent line meets the line y = x, we can set the equations y = -x - 2 and y = x equal to each other:
-x - 2 = x
2x = -2
x = -1
The x-coordinate is -1, and since y = x, the y-coordinate is also -1.
Therefore, the tangent line meets the line y = x at the point (-1, -1).
To find the equation of the tangent to the curve y=2x-x^3 at the point x=-1, we need to find the slope of the curve at that point.
First, let's find the derivative of the curve y=2x-x^3 to find the slope of the tangent line. Taking the derivative of y with respect to x, we get:
dy/dx = d/dx (2x - x^3)
= 2 - 3x^2
Now, substitute x=-1 into the derivative to find the slope at that point:
m = 2 - 3(-1)^2
= 2 - 3(1)
= 2 - 3
= -1
The slope of the tangent line at x=-1 is -1.
Next, we can use the point-slope form of the equation to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1),
where (x1, y1) represents the point on the curve.
Since we are given the point x=-1, we substitute x1 = -1 and y1 = 2(-1) - (-1)^3 = -1.
Therefore, the equation of the tangent line is:
y - (-1) = -1(x - (-1))
y + 1 = -1(x + 1)
y + 1 = -x - 1
y = -x - 2
The equation of the tangent line is y = -x - 2.
To find where the tangent line intersects the line y=x, we can equate the two equations:
-x - 2 = x
Simplifying, we get:
-2x = 2
Dividing by -2, we find:
x = -1
So, the tangent line intersects the line y=x at the point (-1,-1).